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The density of 53.4 wt aqueous NaOH solution is 0.809 g/ml
Given data:
- The mass percent of NaOH is 53.4.
- Volume of NaOH diluted is 16.7 ml.
- The volume of diluted solution is 2.00 L =2000 ml.
- Concentration of diluted solution is 0.169 M.
First, we find the initial concentration of NaOH by using the following formulae,
M₁V₁ = M₂V₂
Where,
M₁ is the initial molarity of NaOH
M₂ is the molarity after dilution
V₁ is the initial volume of NaOH
V₂ is the final volume after dilution.
Substituting the values,
M₁ × 16.7 ml = 0.169 M × 2000 ml,
M₁ =
M₁ = 20.2 M.
Thus, the initial concentration of NaOH is 20.2 M.
we know, 1 M solution contains 1 mol of substance present in 1 L solution,
Thus, 20.2 M solution will have 20.2 mols of NaOH.
Now, we can find the mass of NaOH by using the number of moles and molar mass.
- molar mass of NaOH is 40 g/mol.
Mass = no. of moles × molar mass
= 20.2 mol × 40 g/mol
= 808 g.
Thus, the mass of NaOH is 808g.
53.4 wt of NaOH means 53.4 g of NaOH in a 100 g solution,
Thus, 808 g of NaOH will be present in ,
⇒
⇒ 1513.1 g
Now, Convert the grams of NaOH to milliliters, using the density of NaOH at room temperature.
- The density of NaOH at room temperature is 1.515 g/ml,
Density =
⇒ 1.515 g/mol =
⇒ volume =
⇒ volume = 998.7 ml.
Thus, the volume of NaOH is 998.7 ml.
Hence, we know,
- the mass of NaOH is 808 g
- the volume of NaOH is 998.7 ml
Substituting the values,
Density = 808 g / 998.7 g/ml
⇒ Density = 0.809 g/ml
Thus, the density of 53.4 wt aqueous NaOH is 0.809 g/ml.
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